Question

A standby generator powered by and internal combustion engine has a rated capacity of 100kW. Assuming that the generator is 90% efficient how much horsepower must the engine deliver to the generator at full load? Assuming that the thermal efficiency of the engine is 35% how much diesel fuel will the engine consume in one hour at full load?

Answer 1

Answer:

a)148.9hp

b)25.2kg

Explanation:

Hello!

To solve this exercise follow the steps below.

1. Find the power the generator needs to operate at 100Kw,

This is done by remembering that efficiency in a generator is the ratio between real power and ideal power.

$\alpha g=\frac{Wi}{Wr}$

where

α=generator efficiency =0.9

Wi=nominial or ideal efficiency=100Kw

Wr=real effiency

solving for Wr

$Wr=\frac{Wi}{\alpha g } =\frac{100Kw}{0.9} =111.11kW$

the power that the combustion engine must deliver is 111Kw for the generator to deliver 100Kw, Now use conversion factor to know the value in horsepower

$111.11 kW \frac{1hp}{0.746Kw} =148.9hp$

2. Find the engine supply power using the combustion engine efficiency definition

$\alpha m=\frac{Wo}{Ws}$

α=engine efficiency =0.35

Wo = output power

=111.11Kw

Wi = supply power

solving for Wi

$Wi=\frac{Wo}{\alpha m } =\frac{111.11Kw}{0.35}=317.45Kw$

3. use the calorific power of diessel (CP=45KJ /g) to find the mass flow, remember that the supply power is the product of the mass flow and the caloric power

Wi=m(CP)

m=mass flow

solving for m

$m=\frac{Wi}{CP} =\frac{317.45Kw}{45000KJ/g} =7g/s$

4. The mass flow is the ratio between the mass consumed and the time (1h) in this way we can find the diessel consumed in one hour at full load

M=diessel mass consumed

m=M/t

M=mt

$M=(7 \frac{g}{s})1hour\frac{(3600s)}{1hour} =25200g=25.2Kg$